Suppose you are trying to evaluate the intergral 2x^3dx and you make the substitution u=x^2. The substitution takes one area and converts it to a different one-in this case a trapezoid with vertical sides. Denote by R the region under the first curve from 3 to 3.01 and estimate it's area. Now sketch the region S that the substitution transformed R into and compute its area. Explain how we converted a trall skinny region into a shorter broader one. Especially explain how the du played a role vs. dx.
To evaluate the integral 2x^3 dx using the substitution u = x^2, we can follow these steps:
1. Start by finding the derivative of the substitution u = x^2. Since u = x^2, we can differentiate both sides with respect to x:
du/dx = 2x.
Rearranging the equation, we have dx = du / (2x).
2. Substitute the values of x and dx from the new equation into the original integral. The original integral becomes:
∫(2x^3 dx) = ∫(2x^3) * (du / (2x)).
Simplifying, we get ∫x^2 du.
3. The substitution u = x^2 maps the interval [3, 3.01] in the variable x to the interval [9, 9.0601] in the variable u. This means we need to evaluate the integral ∫(x^2 du) on the interval [9, 9.0601].
4. Now let's consider the region R under the curve from x = 3 to 3.01. This region is a trapezoid with vertical sides, as you mentioned.
To estimate the area of R, we can find the area of a rectangle with width Δx = 0.01 and height f(x), where f(x) is the function 2x^3.
Area of R ≈ Δx * f(x) = 0.01 * (2(3)^3) = 0.54.
So, the estimated area of R is approximately 0.54.
5. The substitution u = x^2 converts the region R into a region S in the u-coordinate system. In this case, S will be a rectangle with width Δu and height f(u), where f(u) is the function x^2 (derived from the substitution).
The width Δu will be the difference between the u-coordinates of the endpoints of R's interval [9, 9.0601], which is given by Δu = 9.0601 - 9 = 0.0601.
The height of S is given by f(u) = x^2 = (u^(1/2))^2 = u.
6. To compute the area of S, multiply the width Δu by the height f(u):
Area of S = Δu * f(u) = 0.0601 * u.
Since u varies from 9 to 9.0601, the area of S can be calculated as:
Area of S = ∫(0.0601*u) du, integrated over the interval [9, 9.0601].
Evaluating this integral, we get:
Area of S = (0.0601/2) * (9.0601^2 - 9^2) = 0.0601 * 0.1201 = 0.00721901.
Therefore, the area of S is approximately 0.00721901.
7. As for the explanation of how the substitution converted a tall skinny region into a shorter broader one:
The original region R was under the curve 2x^3 and ranged from x = 3 to 3.01, which means it was a small strip between these two x-values.
The substitution u = x^2 transformed this strip into a rectangle S with horizontal sides in the u-coordinate system. The substitution allowed us to express the integral in terms of u instead of x, simplifying the evaluation.
The conversion from dx to du played a crucial role in this transformation. When we made the substitution, we altered the differential element from dx to du. In our case, dx was equivalent to du / (2x). Since x was squared in the substitution, it effectively "stretched out" the dx element, making it narrower compared to du. This stretching of the width dx resulted in a wider width du, ultimately transforming the tall skinny region R into the shorter broader region S.